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Difference between revisions of "2007 AMC 12B Problems/Problem 20"
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==Solution 3==
 
==Solution 3==
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Let <math>a</math> and <math>b</math> be the slopes of the lines such that <math>b > a</math> (i.e. the line <math>bx+c</math> is steeper than <math>ax+c</math>) and <math>c > d</math> (i.e. the point <math>(0, c)</math> is higher than the point <math>(0, d)</math>. Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is <math>\frac{bh}{2}</math>, but since a parallelogram is two such triangles, the area becomes <math>bh</math>.
 +
 +
Let <math>b_1</math> and <math>h_1</math> denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and <math>b_2</math> and <math>h_2</math> denote those of the larger parallelogram. Notice that <math>b_1</math> is simple the distance from <math>(0, d)</math> to <math>(0, c)</math>, or <math>(c-d)</math>. Also notice that <math>h_1</math> is the distance from the <math>x</math>-axis to the intersection of lines <math>ax+c</math> and <math>bx+d</math>. This is equivalent to the value of the <math>x</math>-coordinate of intersection, so we solve for <math>x</math>:
 +
 +
 +
<math>ax+c=bx+d</math>
 +
 +
<math>\Rightarrow bx-ax = c-d</math>
 +
 +
<math>\Rightarrow (b-a)x = c-d</math>
 +
 +
<math>\Rightarrow x = \frac{c-d}{b-a}</math>.
 +
 +
 +
The area of the smaller parallelogram is <math>b_1*h_1</math>, or
 +
 +
 +
<math>(c-d) * \frac{c-d}{b-a}</math>
 +
 +
 +
<math>\Rightarrow \frac{(c-d)^2}{b-a}</math>.
 +
 +
 +
<math>b_2</math> is the distance from <math>(0, -d)</math> to <math>(0, c)</math>, or <math>(c+d)</math>. <math>h_2</math> is the <math>x</math>-coordinate of the intersection of the lines <math>ax+c</math> and <math>bx-d</math>. Again, we solve for x:
 +
 +
 +
<math>ax+c=bx-d</math>
 +
 +
<math>\Rightarrow bx-ax = c+d</math>
 +
 +
<math>\Rightarrow (b-a)x = c+d</math>
 +
 +
<math>\Rightarrow x = \frac{c+d}{b-a}</math>.
 +
 +
 +
The area of the larger parallelogram is <math>b-1*h_1</math>, or
 +
 +
 +
<math>(c+d) * \frac{c+d}{b-a}</math>
 +
 +
 +
<math>\Rightarrow \frac{(c+d)^2}{b-a}</math>.
 +
 +
 +
The areas of the parallelograms are given to us: <math>18</math> and <math>72</math>. Therefore we can set up a ratio:
 +
 +
 +
<math>\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}</math>
 +
 +
<math>\Rightarrow 18(c+d)^2 = 72(c-d)^2</math>
  
 
==See also==
 
==See also==

Revision as of 14:56, 22 July 2020

Problem

The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?

$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution

Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which gives $c=3d$ (consider a homothety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$, it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$, so substituting $c = 3d$:

\[\frac{1}{2} (c-d)\left(\frac{c-d}{a-b}\right) = 9 \quad \Longrightarrow \quad 2d^2 = 9(a-b)\]

Thus $3|d$, and we verify that $d = 3$, $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$. Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{\mathbf{(D)} 16}$.

Solution 2

The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$. Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-c}{b-a}$, $\frac{(d-c)^2}{b-a}=18$, and the area contained by the latter is $\frac{(c+d)^2}{b-a}=72$. Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$. $c^2=18(b-a)$ so $c$ is a multiple of $3$. Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$, so the sum is $\boxed{\textbf{(D)} 16}$.

Solution 3

Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$. Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is $\frac{bh}{2}$, but since a parallelogram is two such triangles, the area becomes $bh$.

Let $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$, or $(c-d)$. Also notice that $h_1$ is the distance from the $x$-axis to the intersection of lines $ax+c$ and $bx+d$. This is equivalent to the value of the $x$-coordinate of intersection, so we solve for $x$:


$ax+c=bx+d$

$\Rightarrow bx-ax = c-d$

$\Rightarrow (b-a)x = c-d$

$\Rightarrow x = \frac{c-d}{b-a}$.


The area of the smaller parallelogram is $b_1*h_1$, or


$(c-d) * \frac{c-d}{b-a}$


$\Rightarrow \frac{(c-d)^2}{b-a}$.


$b_2$ is the distance from $(0, -d)$ to $(0, c)$, or $(c+d)$. $h_2$ is the $x$-coordinate of the intersection of the lines $ax+c$ and $bx-d$. Again, we solve for x:


$ax+c=bx-d$

$\Rightarrow bx-ax = c+d$

$\Rightarrow (b-a)x = c+d$

$\Rightarrow x = \frac{c+d}{b-a}$.


The area of the larger parallelogram is $b-1*h_1$, or


$(c+d) * \frac{c+d}{b-a}$


$\Rightarrow \frac{(c+d)^2}{b-a}$.


The areas of the parallelograms are given to us: $18$ and $72$. Therefore we can set up a ratio:


$\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}$

$\Rightarrow 18(c+d)^2 = 72(c-d)^2$

See also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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